Introduction to Complex Analysis
Introduction to Complex Analysis MATH 185
Popular in Course
Popular in Mathematics (M)
This 4 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 185 at University of California - Berkeley taught by Staff in Fall. Since its upload, it has received 54 views. For similar materials see /class/226609/math-185-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.
Reviews for Introduction to Complex Analysis
Report this Material
What is Karma?
Karma is the currency of StudySoup.
Date Created: 10/22/15
Math 185 Solutions to Problems August 30 2006 319 pm Problem 1 Show that complex numbers 2 x ly and special 2by2 matrices Z X y are isomorphic algebraically indistinguishable at least so far as rational arithmetic Ty x operations 7 are concerned Pay close attention to division Solution 1 Obtain Z from Z in the same way as conjugate E is obtained from z namely by reversing the sign of y Then ZT1 Zlzl2 just as lz Elzl2 Consequently every rational operation t sz performed upon complex variables maps turns out to match the analogous operation T WQZ upon special matrices provided WZ is construed as WZ 1 Problem 2 It is tautologically true that Complex variables are unnecessary because anything that can be done with them can also be done without them However this is not the whole truth it omits mention of the extra cost of complications incurred if our thoughts must be formulated in terms solely of real variables with no mention of complex We have already seen that one complex division wz u 1vx 1y u x vyx2 y2 1vx 7 uyx2 y2 entails six real multiplications three real additions and two real divisions these are far more complicated than the concept of complex division Now consider the complex Principal Square Root u 1v w V2 which maps the whole 2 plane ltol onto the righthand half of the wplane including its positive imaginary axis if not all of this axis V2 must be discontinuous as z crosses a Slit along the negative real axis because the inverse map x 1y z w2 takes tiny semicircular neighborhoods around both boundary points ilv of the right half of the wplane to a roughly circular neighborhood around x 7v2 lt 0 in the zplane Given real values for x and y how should u and v be computed using just two not three real square roots and a few rational not transcendental real arithmetic operations and because of the discontinuity comparisons Hint lwl Solution 2 The computed values u and v must satisfy u 2 0 x u2 7 v2 and y 2uv Consequently signy signv when they are nonzero even if u 0 Here is the algorithm u w lxl x2 yz 2 If u0 then vy else if x lt 0 then v CopySigny u u y2v else v y2u CopySigny u here has the magnitude of u but the same sign bit as y has Alas other definitions of CopySign reverse its arguments to remain Compatible with the SIGNu y function provided by ancient Fortran programming languages Note that this algorithm unlike others less carefully designed copes properly with the special cases 2 0 and z x lt 0 The algorithm s use of ixi to first compute u avoids cancellation in case y m 0 gt x Then the test u 0 precludes subsequent division by zero The test x lt 0 compensates for the earlier use of ixi and thus avoids the need to compute a third square root fixi x2 yz 2 which would suffer from roundoff anyway Then the test implicit in the CopySign function implements the complex square root s discontinuity accurately despite roundoff when y is extremely tiny Except across that discontinuity where x lt 0 and y reverses sign this algorithm for x 1y is obviously continuous in x and y A computer implementation must be complicated by measures taken to avoid premature exponent overunderflow when u is first computed Prof W Kahan Page l2 Math 185 Solutions to Problems August 30 2006 319 pm Problem 3 For any nonzero complex constant 9 de ne fz cWzcz Apparently z2 z for all arguments 2 but fz HE for some and fz 7V2 for others which Solution 3 f0 W6 0 Henceforth assume 2 0 let 0 argz and Q argc so 77E lt 0 S T and 77E lt C S TE Now fzZ i1 depending upon 0 and Q not upon 2 nor 9 thus argfzE 072C mod 27E 2 7 072Q 2 mod 27E where in general Hmod27c H if 7TEltHSTE H72TE if TEltHS3TE 0r H27c if 737cltHS77c Consequently H mod 27E2 7 H2 mod 27E 0 if 77E lt H S T 7 if 37cltHS7c 0r 7cltHS37c For our purposes 737 lt H 072Q S 37E whence comes atable showing the ranges of 0 over which fz iWE 0 argz TE lt C argc S 0 0 S Q argc S T fz 7 7x12 2CTE lt 0 s 7 77 lt 0 s 2977 fz 7 NE 77 lt 0 s 29m 2977 lt 0 s 7 Ie fz is a square root of z discontinuous across a slit along argz 2C 7 CopySignQ TE Problem 4 Although for all nonnegative real p and q the principal square root of complex arguments p and q does not always satisfy that identity a For which arguments p and q does q What happens instead b For which complex arguments 2 does Vl 7 22 c For which complex arguments w does w2 7 l Vw7lWwl This problem is very hard if approached naively Solution 4a unless 77E lt argp argq S T Why See Solution 3 Solution 4b Vl 7 22 for all z Why Compare derivatives Solution 4c W2 7 l just when Rew gt 0 0r Rew 0 S Imw Why The neat proof appeals to the Monodromy Theorem which will be taken up later Problem 5 f is the straight line through two given points b and bc in the complex zplane a What equation must 2 satisfy when it lies on f b Find a formula for lz 7 l the shortest distance from z to f if z is not on f c Given a point f not on f lz7fl lz7 l is the equation ofa Parabola whose Focus is f and Direcm39x is Obtain an equivalent equation in terms exclusively of z b c and f without the square root implicit in Solution 5a Imz7bc 0 or equivalently ImEz7b 0 Solution 5b lz7 l lcImz7bcl because the point on f nearest 2 is z 719 Imz7bc Solution 5c z7fcl2 Imz7blt2 or equivalently Ez7f2 ImEz7b2 Solutions for Problems 6 and 7 can be found in the class notes on Conic Sections in the Complex zplane Prof W Kahan Page 22 File Solvef Solution of a Transcendental Equation April 23 2008 519 am A holomorphic function fz is declared to satisfy the equations fz 7 lnl fz 222 f0 0 and f390 1 Here lnl fz is the Principal Logarithm with in S Imagln S TE Other equations satis ed by fz are fz eXpfz 7 222 71 and f39z zl lfz the latter equation is a Singular diiTerential equation with two Regular solutions from which the one selected here satis es the initial conditions f0 0 and f390 1 The selected solution has a Taylor Series expansion about the point z 0 obtainable by repeated implicit differentiation of the equation f f39 2 f z and subsequent substitution of z 0 and f 0 ffquot fi2 2 f 1 1 71 is rejected ffm3fvfuzfu2fv fu023 ffvm4fvfm3fu2zfm3fu fm016 and so on Thus fz z 223 2336 7 24270 254320 2617010 7 139275443200 28204120 7 in the intersection of the domain of fz with the circular disk inside which this series converges What is the radius r of that disk Why is r ZVE rs 35449 Fln The Cauchy Hadamaral formula r lim inf11am la requires a formula for the coef cients an of zn in the series but I know no such formula Numerical computation of the coe icients does not help much because they vary so irregularly that convergence to r is slow for instance i644r144 46945910727y144 3966 WSW45 44171710729y145 4267 la46l 146 33217610728y146 3957 marl47 13148210728y147 3919 la48l 144 25996310729 148 3940 The series radius of convergence is the distance r from z 0 to the nearest singularity of a function Fz that agrees with fz on the intersection of their domains but the domain of Fz extends as far as analytic continuation can take it beyond the domain S of fz which is the region swept out by z i 2w ilnlw as w sweeps through the whole complex plane Domain S 7S g lies inside a fourpointed star the image of logarithm s slit where w lt 71 Dumam ui fz Prof W Kahan Math 185 Page 12 File Solvef Solution of a Transcendental Equation April 23 2008 519 am If 2 i 2w7lnlw is to cross from inside S to outside this starshaped region w must cross the logarithm s slit crossing the line where w lt 7l and lnlw jumps by i217 but for the sake of 2 s continuity lnlw must be replaced by an adjacent branch of the logarithm function In other words only for n i1 can 2 i 2w 7lnlw 7 2n17c get somewhat beyond the boundary of S into a larger domain for the analytic continuation Fz of fz Within this tentatively extended domain Fz satis es Fz 7 lnl Fz i 217E 222 and Fz exp Fz 7 222 7 l and F39z z 1 lFz The domain is extended tentatively because it contains singularities of Fz that affect its analytic continuation beyond them The tentative enlargement lies between the star g shown above and a bigger star whose boundary is traced by z i 2w 7lnlw i 2170 as w runs along the slit whereon w lt 7l Here is a picture with singularities marked by x I v I I I I I I I I I mz O I I quot35i3392391396139 i5 Rez Fz has singularities inside the tentatively extended domain F39z oo wherever Fz 0 i z The singularities nearest 2 0 are Branch Points at i2 ZVE linoE rs 25066li1 These determine the radius of convergence of the Taylor series it is r ZVE As 2 traces a path starting at z 0 Fz s value depends on how the path wends among the singularities This problem illustrates an obstacle impeding the automation of the algebra of transcendental complex analytic functions Computerized algebra systems like MAPLE MATHEMATICA and DERIVE can cope with only the simplest domains if any For instance these programs can compute as many as you like of the coef cients of the Taylor series of Fz but not r The accurate numerical computation of Fz is another interesting story for some other day Prof W Kahan Math 185 Page 22